What is the extreme f(x)=20sin(2x) ?

The extreme f(x)=20sin(2x) is Maximum(pi/4+pin,20),Minimum((3pi)/4+pin,-20)

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extreme f(x)=20sin(2x)

extreme points

f (x) = 20 sin (2 x)

Maximum (\frac{π}{4} + πn, 20), Minimum (\frac{3 π}{4} + πn, - 20)

Solution steps

Find the critical points:

x = \frac{π}{4} + πn, x = \frac{3 π}{4} + πn

Domain of

20 sin (2 x) : - \infty < x < \infty

Find intervals:

Increasing

: πn \leq x < \frac{π}{4} + πn,

Decreasing

: \frac{π}{4} + πn < x < \frac{3 π}{4} + πn,

Increasing

: \frac{3 π}{4} + πn < x < π + πn

Plug the extreme point

x = \frac{π}{4} + πn

into

20 sin (2 x) \Rightarrow y = 20

Maximum (\frac{π}{4} + πn, 20)

Plug the extreme point

x = \frac{3 π}{4} + πn

into

20 sin (2 x) \Rightarrow y = - 20

Minimum (\frac{3 π}{4} + πn, - 20)

Maximum (\frac{π}{4} + πn, 20), Minimum (\frac{3 π}{4} + πn, - 20)

x (3 - x)

f (x) = y \cdot e^{x} + x \cdot ln (y)

f (x) = (x^{(\frac{2}{3})}) (1 - x^{2})

f (x, y) = \frac{69120}{x} + \frac{69120}{y} + 5 x y

f (x) = 4 x^{2} - 8 x + 7 y^{2} + 8

What is the extreme f(x)=20sin(2x) ?
The extreme f(x)=20sin(2x) is Maximum(pi/4+pin,20),Minimum((3pi)/4+pin,-20)